Differentiating imaginary numbers

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  1. Occams Chainsaw's Avatar
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    Differentiating imaginary numbers
    Is it possible to differentiate i?
    I can't work out what I would do with \sqrt {-1} to get it to differentiate. Any help?
  2. Slumpy's Avatar
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    Re: Differentiating imaginary numbers
    i is a constant.
  3. Megzx13's Avatar
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    Re: Differentiating imaginary numbers
    dont think you can, as minus square roots can't be changed into an indices? I think :P
  4. bananarama2's Avatar
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    Re: Differentiating imaginary numbers
    It's a constant?
  5. Occams Chainsaw's Avatar
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    Re: Differentiating imaginary numbers
    So far I have got that
    \sqrt {-1} = -1^{1/2} = i



dy/di x di/dx



(dy^{1/2}\div di)(di\div dx)

    so now I need to differentiate in respects to i right?
    Last edited by Occams Chainsaw; 26-09-2012 at 18:34.
  6. poohat's Avatar
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    Re: Differentiating imaginary numbers
    as Slumpy said, i is a constant. However functions of complex variables can be differentiated. This is done by differentiation the real and imaginary parts seperately and using the Cauchy Riemann equations, see: http://en.wikipedia.org/wiki/Cauchy%...mann_equations (its unlikely that page will be understandable if you havent taken at least undergrad real analysis, but you can google for 'complex differentiation' and try to find something more elementary)
  7. bananarama2's Avatar
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    Re: Differentiating imaginary numbers
    (Original post by Occams Chainsaw)
    So far I have got that
    \sqrt {-1} = -1^1/2 = i



dy/di x di/dx



(dy^(1/2)/di)(di/dx)
    Does  \frac{dy}{d2} make sense?

    You can differentiate complex variables though.


     f'(z_0 ) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}

    Etc.... http://www.ams.org/bookstore/pspdf/mbk-49-prev.pdf
  8. poohat's Avatar
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    Re: Differentiating imaginary numbers
    for example : http://www.ams.org/bookstore/pspdf/mbk-49-prev.pdf (although this still presupposes a basic background in analysis)
  9. Occams Chainsaw's Avatar
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    Re: Differentiating imaginary numbers
    (Original post by poohat)
    as Slumpy said, i is a constant. However functions of complex variables can be differentiated. This is done by differentiation the real and imaginary parts seperately and using the Cauchy Riemann equations, see: http://en.wikipedia.org/wiki/Cauchy%...mann_equations (its unlikely that page will be understandable if you havent taken at least undergrad real analysis, but you can google for 'complex differentiation' and try to find something more elementary)
    Thanks
  10. bananarama2's Avatar
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    Re: Differentiating imaginary numbers
    (Original post by bananarama2)
    Does  \frac{dy}{d2} make sense?

    You can differentiate complex variables though.


     f'(z_0 ) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}

    Etc.... http://www.ams.org/bookstore/pspdf/mbk-49-prev.pdf

    (Original post by poohat)
    for example : http://www.ams.org/bookstore/pspdf/mbk-49-prev.pdf (although this still presupposes a basic background in analysis)

    Snap
  11. Slumpy's Avatar
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    Re: Differentiating imaginary numbers
    (Original post by Occams Chainsaw)
    So far I have got that
    \sqrt {-1} = -1^{1/2} = i



dy/di x di/dx



(dy^{1/2}\div di)(di\div dx)

    so now I need to differentiate in respects to i right?
    Ignoring the fact I already told you i is a constant, what are you trying to do here? You've gone from the definition of i to a chain rule version of dy/dx, without having defined a y or x. Then your next line is a bit weird.
  12. Occams Chainsaw's Avatar
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    Re: Differentiating imaginary numbers
    (Original post by Slumpy)
    Ignoring the fact I already told you i is a constant, what are you trying to do here? You've gone from the definition of i to a chain rule version of dy/dx, without having defined a y or x. Then your next line is a bit weird.
    Your right.

    I have just started c1 maths in college but I have been playing around with further maths and certain constants such as i and e and I guess my lack of understanding showed there Thanks for your help anyway.
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